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Nastasia [14]
4 years ago
11

Find the expression that is equilvelent

Mathematics
2 answers:
Elodia [21]4 years ago
4 0
For this case we have the following expression:
 -18a ^ -2b ^ 5 / -12a ^ -4b ^ -6
 For power properties we have:
 -18a ^ (- 2 - (- 4)) b ^ (5 - (- 6)) / - 12
 Rewriting we have:
 18a ^ (- 2 + 4) b ^ (5 + 6) / 12
 3a ^ (- 2 + 4) b ^ (5 + 6) / 2
 3a ^ 2b11 / 2
 Answer:
 
3a ^ 2b11 / 2
 
option 3
Ganezh [65]4 years ago
3 0
\dfrac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\dfrac{18:6}{12:6}\cdot a^{-2-(-4)}b^{5-(-6)}=\dfrac{3}{2}\cdot a^{-2+4}b^{5+6}=\dfrac{3a^2b^{11}}{2}


Used:\\\\a^n:a^m=\dfrac{a^n}{a^m}=a^{n-m}
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Every day, Katie looks for seashells on the beach. She has 282828 shells in her collection. Katie finds 121212 more shells each
trasher [3.6K]

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4 0
4 years ago
Please I want the area and perimeter of this shape!!<br> With steps and explanation please!!
worty [1.4K]

For perimeter you will add each side together so

8ft+22ft+12ft+16ft=58

So your perimeter is 58ft.

For area your going to divide the shape into 2 to get the area as this is an irregular shape, so

Base 1+base2x height / by 2

22+8 x4/2

30x4/2

120/2

60

If you’re wondering how I got 4 I subtracted 16 with 12 to know what the height was. Good luck!!

7 0
3 years ago
Please answer the questions
siniylev [52]

Answer:

  1. root (7²+8²)=10.63
  2. root (10²-5²)=8.66
  3. root (11²-3²)=10.58
  4. root (18²+15²)=23.43

Step-by-step explanation:

a²+b² = c²

a = side of right triangle

b = side of right triangle

c = hypotenuse

6 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Show all work to receive credit.
Helga [31]

Answer:

angle 1= 70 degrees

angle 2= 65 degrees

angle 3= 65 degrees

angle 4 = 115 degrees

angle 5= 45 degrees

angle 6= 45 degrees

angle 7= 45 degrees

Step-by-step explanation:

8 0
4 years ago
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